# Question on gravitation?

## Two small spaceships, each with mass m = 2000 kg, are in the circular Earth orbit of the figure, at an altitude h of 440 km. Igor, the commander of one of the ships, arrives at any fixed point in the orbit 84.0 s ahead of Picard, the commander of the other ship. What are the (a) period T0 and (b) speed v0 of the ships? At point P in the figure, Picard fires an instantaneous burst in the forward direction, reducing his ship's speed by 1.30%. After this burst, he follows the elliptical orbit shown dashed in the figure. What are the (c) kinetic energy and (d) potential energy of his ship immediately after the burst? In Picard’s new elliptical orbit, what are (e) the total energy E, (f) the semimajor axis a, and (g) the orbital period T? (h) How much earlier than Igor will Picard return to P? Aug 1, 2018

(a) For object of mass $m = 2000 \setminus k g$ moving in a circular orbit of radius $r$ with a speed ${v}_{0}$ around the earth of mass $M$ (at an altitude $h$ of $440 \setminus m$), the orbital period ${T}_{0}$ is given by Kepler's third law.

${T}_{0}^{2} = \frac{4 {\pi}^{2}}{G M} {r}^{3}$ ......(1)
where $G$ is Universal Gravitational Constant.

In terms of altitude of the spaceships

${T}_{0} = \sqrt{\frac{4 {\pi}^{2}}{G M} {\left(R + h\right)}^{3}}$

Inserting various values we get
${T}_{0} = \sqrt{\frac{4 {\pi}^{2}}{\left(6.67 \times {10}^{-} 11\right) \left(5.98 \times {10}^{24}\right)} {\left(6.37 \times {10}^{6} + 4.40 \times {10}^{5}\right)}^{3}}$
$\implies {T}_{0} = \sqrt{\frac{4 {\pi}^{2}}{\left(6.67 \times {10}^{-} 11\right) \left(5.98 \times {10}^{24}\right)} {\left(6.81 \times {10}^{6}\right)}^{3}}$
$\implies {T}_{0} = \sqrt{\frac{4 {\pi}^{2}}{\left(6.67 \times {10}^{-} 11\right) \left(5.98 \times {10}^{24}\right)} {\left(6.81 \times {10}^{6}\right)}^{3}}$
$\implies {T}_{0} = 5591.0 \setminus s$

(b) Centripetal force is balanced by gravitational force. The expression becomes

$\frac{m {v}_{0}^{2}}{r} = \frac{G M m}{r} ^ 2$
$\implies {v}_{0} = \sqrt{\frac{G M}{r}}$

Alternately, for a circular orbit

${v}_{0} = r \omega$
$\implies {v}_{0} = \left(R + h\right) \frac{2 \pi}{T} _ 0$

Inserting various values in alternate expression

${v}_{0} = \left(6.81 \times {10}^{6}\right) \frac{2 \pi}{5591}$
$\implies {v}_{0} = 7653 \setminus m \cdot {s}^{-} 1$

(c) Kinetic energy of Picard's spaceship just after firing of burst

${E}_{K} = \frac{1}{2} m {v}^{2}$

Inserting various values we get

${E}_{K} = \frac{1}{2} \left(2000\right) {\left(\frac{100 - 1.30}{100} \times 7653\right)}^{2}$
$\implies {E}_{K} = 5.7 \times {10}^{10} \setminus J$

(d) Potential energy of this spaceship at the same time

${E}_{P} = - \frac{G M m}{r}$

Inserting various values we get

${E}_{P} = - \frac{\left(6.67 \times {10}^{-} 11\right) \left(5.98 \times {10}^{24}\right) \left(2000\right)}{6.81 \times {10}^{6}}$
${E}_{P} = - 1.17 \times {10}^{11} \setminus J$

(e) Total energy ${E}_{T} = {E}_{P} + {E}_{K}$

${E}_{T} = - 1.17 \times {10}^{11} + 5.7 \times {10}^{10}$
${E}_{T} = - 6.0 \times {10}^{10} \setminus J$

(f) Semi major $a$ axis is given by

${E}_{T} = - \frac{G M m}{2 a}$
$\implies a = - \frac{G M m}{2 {E}_{T}}$

Inserting given values we get

$\implies a = - \frac{\left(6.67 \times {10}^{-} 11\right) \left(5.98 \times {10}^{24}\right) \left(2000\right)}{2 \left(- 6.0 \times {10}^{10}\right)}$
$\implies a = 6.65 \times {10}^{6} \setminus m$

(g) The new orbital period $T$ is found from the expression

${T}^{2} = \frac{4 {\pi}^{2}}{\left(6.67 \times {10}^{-} 11\right) \left(5.98 \times {10}^{24}\right)} {\left(6.65 \times {10}^{6}\right)}^{3}$
$\implies T = \sqrt{\frac{4 {\pi}^{2}}{\left(6.67 \times {10}^{-} 11\right) \left(5.98 \times {10}^{24}\right)} {\left(6.65 \times {10}^{6}\right)}^{3}}$
$\implies T = 5395.1 \setminus s$

(h) Picard is now faster than Igor by time

$\Delta T = 5591.0 - 5395.1 = 195.9 \setminus s$

When he arrived first at point $P$ he was $84.0 \setminus s$ behind. Now earlier by

$195.9 - 84.0 = 111.9 \setminus s$